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--- virasoro_symmetries [2021/09/15 17:31]
Brian McPeak
+++ virasoro_symmetries [2021/09/15 17:52] (current)
Brian McPeak
@@ Line 68: / Line 68: @@
 The first are essentially those where every element of the ordered basis $\left\{ L_{-n_1} ... L_{-n_p} \right\}_{n_1 \leq n_2 ... \leq n_p >0}$ creates a linearly independent element when acting on $|\Delta \rangle$. Such representations are called **Verma modules**, denoted by $\mathcal{V}_\Delta$. Verma modules are infinite-dimensional representations, and are isomorphic to $U(\mathcal{V^+})$.
-====Singular vectors====
+===Null vectors===
-Consider now a subrepresentation of $\mathcal{V}_\Delta$. It must also have $L_0$ eigenvalues bounded from below, and have a highest-weight vector which we denote $|\chi \rangle$. For a non-trivial subrepresentation, $|\chi \rangle \neq |\Delta \rangle$, so $|\chi \rangle$ must be a descendent in the representation $\mathcal{V}_\Delta$. We conclude that $\mathcal{V}_\Delta$ has subrepresentations when there exists a descendent $|\chi \rangle$ which is also annihilated by all $L_n$ for $n>0$. The state $|\chi \rangle$ is called a **singular vector** or a **null vector**.
+Another possibility is to create a finite-dimensional representation by quotienting a Verma module by one of its subrepresentations.  Consider a subrepresentation of $\mathcal{V}_\Delta$. It must also have $L_0$ eigenvalues bounded from below, and have a highest-weight vector which we denote $|\chi \rangle$. For a non-trivial subrepresentation, $|\chi \rangle \neq |\Delta \rangle$, so $|\chi \rangle$ must be a descendent in the representation $\mathcal{V}_\Delta$. We conclude that $\mathcal{V}_\Delta$ has subrepresentations when there exists a descendent $|\chi \rangle$ which is also annihilated by all $L_n$ for $n>0$. The state $|\chi \rangle$ is called a **null vector** or a **singular vector**.
 ==Example: null vectors at level 1==
+The only descendent at level 1 is $|\chi \rangle = L_{-1} | \Delta \rangle$. We now need to check that $L_n$ annihilates $|\chi \rangle $ for all $n \geq 1$. This is automatic for  $n \geq 2$ because $|\Delta \rangle$ is a highest-weight state.
+So we need only check $L_1$.
+$$ L_1 |\chi \rangle = L_1 L_{-1} |\Delta \rangle = 2 \Delta |\Delta \rangle \, . $$
+Therefore $\mathcal{V}_\Delta$ has a null vector if and only if $\Delta = 0$.
 ==Example: null vectors at level 2==
-===Denegerate representations===
+We can do the same thing at level two. Now there is a family of possible null vectors
-Another possibility is to create a finite-dimensional representation by quotienting a Verma module by one of its subrepresentations. Consider a Verma module $\mathcal{V}_\Delta$. It's $L_0$ eigenvalues are bounded from below. Therefore the $L_0$ eigenvalues of its subrepresentations are also bounded from below, so its subrepresentations also from
+$$| \chi \rangle = \left( a L_{-1}^2 + b L_{-2} \right) |\Delta \rangle \, . $$
+We need to check that $L_1 |\chi \rangle = 0$ and $L_2 |\chi \rangle = 0$. These equations can only be solved when
+$$ \Delta = \frac{5 - c \pm \sqrt{(c - 1)(c - 25)}}{16} \, . $$
+In particular, we only find level 2 null vectors when $c \leq 1$ or $c \geq 25$.
+===Denegerate representations===
+If a Verma module $\mathcal{V}_\Delta$ has null vectors, then we can construct a new highest-weight representation by quotienting it by the subrepresentation constructed from its null vectors. in other words,
-This forms a **degenerate representation**.  All representations fall into one of these two classes.
+$$ \mathcal R = \frac{\mathcal{V}_\Delta}{U(\mathcal{V}^+) |\chi \rangle} \, . $$
+This forms a **degenerate representation**.  All representations are either Verma modules or degenerate representations. The latter are important in the study of [[minimal models]].

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